{"id":229571,"date":"2025-06-08T09:40:38","date_gmt":"2025-06-08T09:40:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229571"},"modified":"2025-06-08T09:40:40","modified_gmt":"2025-06-08T09:40:40","slug":"propose-an-efficient-synthesis-of-propanoic-acid-from-acetylene-cooh-your-answer-incorrect","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/propose-an-efficient-synthesis-of-propanoic-acid-from-acetylene-cooh-your-answer-incorrect\/","title":{"rendered":"Propose an efficient synthesis of propanoic acid from acetylene: COOH Your answer incorrect."},"content":{"rendered":"\n

Propose an efficient synthesis of propanoic acid from acetylene: COOH Your answer incorrect. Try again. Draw the structure of X, HC\u00e2\u2030\u00a1CH Edit Your answer correct. Which reagents will accomplish the transformation shown above? A. 1) NaNH2 2) CH3CH2OH B. O2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To efficiently synthesize propanoic acid<\/strong> (CH\u2083CH\u2082COOH) from acetylene<\/strong> (HC\u2261CH), we need to perform a series of reactions that build a 3-carbon carboxylic acid from a 2-carbon alkyne. Let\u2019s walk through the correct synthetic route:<\/p>\n\n\n\n

\n\n\n\n

Step-by-step Reaction Pathway:<\/h3>\n\n\n\n
    \n
  1. Start with acetylene (HC\u2261CH)<\/strong>
    Acetylene has two sp-hybridized carbons and serves as a 2-carbon building block.<\/li>\n\n\n\n
  2. Step A: Alkylation with ethyl bromide (CH\u2083CH\u2082Br)<\/strong>
    Use NaNH\u2082<\/strong> to deprotonate acetylene, forming the acetylide anion (HC\u2261C\u207b). Then add ethyl bromide (CH\u2083CH\u2082Br)<\/strong> to perform nucleophilic substitution<\/strong>, extending the carbon chain: HC\u2261CH\u2192CH\u2083CH\u2082Br1.NaNH2HC\u2261CCH2CH3HC\u2261CH \\xrightarrow[\\text{CH\u2083CH\u2082Br}]{1. NaNH\u2082} HC\u2261CCH\u2082CH\u2083HC\u2261CH1.NaNH2\u200bCH\u2083CH\u2082Br\u200bHC\u2261CCH2\u200bCH3\u200b This gives 1-butyne<\/strong>.<\/li>\n\n\n\n
  3. Step B: Oxidative cleavage to carboxylic acid<\/strong>
    Oxidize 1-butyne with O\u2083 (ozone)<\/strong> followed by oxidative work-up (e.g., H\u2082O\u2082)<\/strong> to cleave the triple bond and form carboxylic acids: HC\u2261CCH2CH3\u21921.\u00a0O\u2083,\u00a02.\u00a0H\u2082O\u2082CH3CH2COOHHC\u2261CCH\u2082CH\u2083 \\xrightarrow{\\text{1. O\u2083, 2. H\u2082O\u2082}} CH\u2083CH\u2082COOHHC\u2261CCH2\u200bCH3\u200b1.\u00a0O\u2083,\u00a02.\u00a0H\u2082O\u2082\u200bCH3\u200bCH2\u200bCOOH This yields propanoic acid<\/strong> as the final product.<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Structure of Intermediate X:<\/h3>\n\n\n\n

    The intermediate X<\/strong> is HC\u2261CCH\u2082CH\u2083<\/strong> (1-butyne), not ethene (H\u2082C=CH\u2082<\/strong>) as shown in the incorrect student response.<\/p>\n\n\n\n

    \n\n\n\n

    Explanation <\/h3>\n\n\n\n

    The transformation of acetylene into propanoic acid is a classic carbon-chain extension followed by oxidation. Acetylene (HC\u2261CH) is a highly acidic alkyne, and treatment with a strong base like NaNH\u2082<\/strong> generates an acetylide anion. This nucleophilic species is ideal for SN2 alkylation<\/strong> with a primary alkyl halide such as ethyl bromide (CH\u2083CH\u2082Br), forming 1-butyne (HC\u2261CCH\u2082CH\u2083).<\/p>\n\n\n\n

    The next critical step is the oxidative cleavage<\/strong> of the carbon-carbon triple bond. Ozonolysis, particularly with a peroxide work-up (like H\u2082O\u2082), cleaves alkynes into two carboxylic acid groups. Since the triple bond is internal (not terminal), the reaction produces two acids. For 1-butyne, cleavage gives propanoic acid<\/strong> and formic acid<\/strong>, but under conditions aimed specifically at the terminal portion, propanoic acid is the major product of interest.<\/p>\n\n\n\n

    The incorrect structure of X (ethene, H\u2082C=CH\u2082) indicates a misunderstanding\u2014perhaps confusing a double bond product instead of the required triple-bonded intermediate. Ethene cannot be converted directly into propanoic acid via this method.<\/p>\n\n\n\n

    Thus, the correct reagent sequence is:<\/p>\n\n\n\n

      \n
    • A.<\/strong> 1) NaNH\u2082 2) CH\u2083CH\u2082Br<\/li>\n\n\n\n
    • B.<\/strong> O\u2083\/H\u2082O\u2082<\/li>\n<\/ul>\n\n\n\n

      This method efficiently builds the 3-carbon chain and oxidizes it to the desired propanoic acid<\/strong>.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Propose an efficient synthesis of propanoic acid from acetylene: COOH Your answer incorrect. Try again. Draw the structure of X, HC\u00e2\u2030\u00a1CH Edit Your answer correct. Which reagents will accomplish the transformation shown above? A. 1) NaNH2 2) CH3CH2OH B. O2 The Correct Answer and Explanation is: To efficiently synthesize propanoic acid (CH\u2083CH\u2082COOH) from acetylene (HC\u2261CH), […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-229571","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229571","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=229571"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229571\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229571"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229571"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}