{"id":87021,"date":"2025-06-01T04:26:24","date_gmt":"2025-06-01T04:26:24","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87021"},"modified":"2025-06-01T04:26:24","modified_gmt":"2025-06-01T04:26:24","slug":"the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/","title":{"rendered":"The reaction of 1,2-dibromoethane with alcoholic KOH yields"},"content":{"rendered":"

The reaction of 1,2-dibromoethane with alcoholic KOH yields<\/p>\n

[amp_mcq option1=”ethene” option2=”ethyne” option3=”1-bromo-2-hydroxyethane” option4=”1-bromoethene” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\n1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene).
\n<\/section>\n
\nReaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.
\n<\/section>\n
\nStep 1: CH\u2082Br-CH\u2082Br + KOH (alc.) \u2192 CH\u2082=CHBr + KBr + H\u2082O
\nStep 2: CH\u2082=CHBr + KOH (alc.) \u2192 HC\u2261CH + KBr + H\u2082O
\nThe final product is ethyne. Ethene would result from single dehydrohalogenation of a mono-halogenated alkane or dehalogenation of a vicinal dihalide using a different reagent (e.g., Zn dust). 1-bromo-2-hydroxyethane would be a substitution product, which is less favored with alcoholic KOH, which promotes elimination. 1-bromoethene is an intermediate in the reaction, not the final product under excess alcoholic KOH.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The reaction of 1,2-dibromoethane with alcoholic KOH yields [amp_mcq option1=”ethene” option2=”ethyne” option3=”1-bromo-2-hydroxyethane” option4=”1-bromoethene” correct=”option2″] This question was previously asked in UPSC Geoscientist – 2023 Download PDFAttempt Online 1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, … <\/p>\n

Detailed SolutionThe reaction of 1,2-dibromoethane with alcoholic KOH yields<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1091],"tags":[1105,1096,1272],"class_list":["post-87021","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1105","tag-chemistry","tag-organic-chemistry","no-featured-image-padding"],"yoast_head":"\nThe reaction of 1,2-dibromoethane with alcoholic KOH yields<\/title>\n<meta name=\"description\" content=\"1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene). Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The reaction of 1,2-dibromoethane with alcoholic KOH yields\" \/>\n<meta property=\"og:description\" content=\"1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene). Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:26:24+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"The reaction of 1,2-dibromoethane with alcoholic KOH yields","description":"1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene). Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/","og_locale":"en_US","og_type":"article","og_title":"The reaction of 1,2-dibromoethane with alcoholic KOH yields","og_description":"1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene). Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:26:24+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/","name":"The reaction of 1,2-dibromoethane with alcoholic KOH yields","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:26:24+00:00","dateModified":"2025-06-01T04:26:24+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"1,2-dibromoethane is a vicinal dibromide (bromine atoms on adjacent carbons). Alcoholic KOH is a strong base that promotes elimination reactions (dehydrohalogenation). When 1,2-dibromoethane is treated with alcoholic KOH, a double dehydrohalogenation occurs. First, one molecule of HBr is eliminated to form 1-bromoethene (vinyl bromide). Then, a second molecule of HBr is eliminated from 1-bromoethene to form ethyne (acetylene). Reaction of a vicinal dihalide with alcoholic KOH results in double dehydrohalogenation, yielding an alkyne.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-reaction-of-12-dibromoethane-with-alcoholic-koh-yields\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"The reaction of 1,2-dibromoethane with alcoholic KOH yields"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87021","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87021"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87021\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87021"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87021"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87021"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}