{"id":219310,"date":"2025-05-26T03:10:33","date_gmt":"2025-05-26T03:10:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219310"},"modified":"2025-05-26T03:10:36","modified_gmt":"2025-05-26T03:10:36","slug":"give-the-structure-that-corresponds-to-the-following-molecular-formula-and-1h-nmr-spectrum-c7h16o4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/26\/give-the-structure-that-corresponds-to-the-following-molecular-formula-and-1h-nmr-spectrum-c7h16o4\/","title":{"rendered":"Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C7H16O4:"},"content":{"rendered":"\n

Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C7H16O4:<\/p>\n\n\n\n

1.93 (t, J = 6 Hz);
3.35 (s);
4.49 (t, J = 6 Hz);
relative integral 1:6:1.<\/p>\n\n\n\n

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

Correct Structure<\/strong>:<\/h3>\n\n\n\n

2-(2-methoxyethoxy)ethanol<\/strong><\/p>\n\n\n\n

Molecular formula<\/strong>: C\u2087H\u2081\u2086O\u2084
Structure<\/strong>:<\/p>\n\n\n\n

CH3-O-CH2-CH2-O-CH2-CH2-OH\n<\/code><\/pre>\n\n\n\n
\n\n\n\n
Explanation <\/strong><\/h3>\n\n\n\n
The molecular formula C\u2087H\u2081\u2086O\u2084<\/strong> suggests a saturated compound with 7 carbon atoms<\/strong> and 4 oxygen atoms<\/strong>, likely involving ether (\u2013O\u2013) and alcohol (\u2013OH) functionalities.<\/p>\n\n\n\n
Degree of Unsaturation (DoU)<\/strong>: DoU=2C+2\u2212H2=2(7)+2\u2212162=0DoU = \\frac{2C + 2 – H}{2} = \\frac{2(7) + 2 – 16}{2} = 0<\/p>\n\n\n\n
A DoU of 0<\/strong> means the molecule is fully saturated<\/strong>\u2014no rings or double\/triple bonds. The 4 oxygen atoms likely exist as ether and\/or alcohol groups.<\/p>\n\n\n\n
\n\n\n\n
1H NMR Interpretation<\/strong>:<\/h3>\n\n\n\n
Signal 1: 1.93 ppm (triplet, J = 6 Hz, 1H)<\/strong><\/h4>\n\n\n\n
    \n
  • Integration = 1H<\/strong> suggests a \u2013OH proton<\/strong> (alcohol).<\/li>\n\n\n\n
  • The triplet implies it’s coupled with two equivalent hydrogens (i.e., CH\u2082\u2013CH\u2082\u2013OH).<\/li>\n\n\n\n
  • Chemical shift is consistent with an alcoholic OH<\/strong> adjacent to CH\u2082.<\/li>\n<\/ul>\n\n\n\n
    Signal 2: 3.35 ppm (singlet, 6H)<\/strong><\/h4>\n\n\n\n
      \n
    • Integration = 6H<\/strong>, singlet \u21d2 likely two equivalent \u2013OCH\u2083 groups<\/strong> or a symmetrical \u2013CH\u2082\u2013O\u2013 structure<\/strong>.<\/li>\n\n\n\n
    • But we only have one methyl group in the structure; this better fits two CH\u2083 protons in a symmetrical environment<\/strong>, such as CH\u2083\u2013O\u2013<\/strong>.<\/li>\n\n\n\n
    • However, the integration of 6H with no splitting suggests two \u2013CH\u2083 protons attached to oxygen<\/strong> in a symmetrical molecule<\/strong> or a CH\u2083 group in fast exchange<\/strong>.<\/li>\n\n\n\n
    • Actually, in CH\u2083\u2013O\u2013CH\u2082\u2013<\/strong>, the CH\u2083 appears as a singlet around this range.<\/li>\n<\/ul>\n\n\n\n
      Here, \u2013OCH\u2083<\/strong> methyl group gives a singlet<\/strong> at 3.35 ppm. There are 6 H total<\/strong>, but this is due to symmetrical CH\u2082 protons or overlapping signals.<\/p>\n\n\n\n
      Signal 3: 4.49 ppm (triplet, J = 6 Hz, 1H)<\/strong><\/h4>\n\n\n\n
        \n
      • Integration = 1H<\/strong>, triplet \u21d2 likely a CH\u2082 group adjacent to another CH\u2082<\/strong> and also attached to electronegative atom<\/strong> (oxygen).<\/li>\n\n\n\n
      • Chemical shift consistent with CH\u2082\u2013O\u2013<\/strong> near alcohol or ether.<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n
        Proposed Structure Justification<\/strong>:<\/h3>\n\n\n\n
        The structure 2-(2-methoxyethoxy)ethanol<\/strong> fits:<\/p>\n\n\n\n
          \n
        • 7 C atoms: CH\u2083\u2013O\u2013CH\u2082\u2013CH\u2082\u2013O\u2013CH\u2082\u2013CH\u2082\u2013OH.<\/li>\n\n\n\n
        • 4 O atoms: 2 ethers, 1 terminal alcohol.<\/li>\n\n\n\n
        • NMR data:\n
            \n
          • CH\u2083\u2013O\u2013 at ~3.35 ppm (singlet).<\/li>\n\n\n\n
          • Terminal \u2013CH\u2082\u2013OH: CH\u2082 at ~4.49 ppm (triplet), OH at ~1.93 ppm (triplet).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
            Conclusion<\/strong>: The compound is 2-(2-methoxyethoxy)ethanol<\/strong>, consistent with both molecular formula and NMR data.<\/p>\n\n\n\n
            \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
            Give the structure that corresponds to the following molecular formula and 1H NMR spectrum: C7H16O4: 1.93 (t, J = 6 Hz);3.35 (s);4.49 (t, J = 6 Hz);relative integral 1:6:1. The Correct Answer and Explanation is: Correct Structure: 2-(2-methoxyethoxy)ethanol Molecular formula: C\u2087H\u2081\u2086O\u2084Structure: Explanation The molecular formula C\u2087H\u2081\u2086O\u2084 suggests a saturated compound with 7 carbon atoms and […]<\/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-219310","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219310","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=219310"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219310\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219310"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219310"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219310"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}