{"id":27393,"date":"2025-06-19T22:40:49","date_gmt":"2025-06-19T22:40:49","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=27393"},"modified":"2025-06-19T22:40:50","modified_gmt":"2025-06-19T22:40:50","slug":"sodium-borodeuteride-nabd4-is-a-reagent-that-has-analogous-reactivity-to-sodium-borohydride","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/sodium-borodeuteride-nabd4-is-a-reagent-that-has-analogous-reactivity-to-sodium-borohydride\/","title":{"rendered":"Sodium borodeuteride (NaBD4) is a reagent that has analogous reactivity to sodium borohydride."},"content":{"rendered":"\n

Sodium borodeuteride (NaBD4) is a reagent that has analogous reactivity to sodium borohydride. It acts as a source of deuteride (D-) instead of hydride (H-). Which of the structures below (a-f) is the product of the following reaction? NaBD4 + CH3OH OH OH OH HO- OH H2CO OCH3<\/p>\n\n\n\n

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

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

The correct answer is b<\/strong>.<\/p>\n\n\n\n

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

The starting material shown in the reaction is 2,3-butanedione<\/strong> (also known as diacetyl), which is a diketone with the structure:<\/p>\n\n\n\n

CopyEditCH3\u2013CO\u2013CO\u2013CH3\n<\/code><\/pre>\n\n\n\n
In the reaction, this compound is treated with NaBD4<\/strong> in methanol (CH3OH)<\/strong>. Sodium borodeuteride behaves like sodium borohydride (NaBH4), but instead of donating hydride ions (H\u207b), it donates deuteride ions (D\u207b)<\/strong>.<\/p>\n\n\n\n
Reduction of Ketones:<\/strong><\/p>\n\n\n\n
NaBD4 reduces ketones to secondary alcohols<\/strong> by delivering a deuteride to the carbon of the carbonyl group (C=O), breaking the double bond and converting it to a hydroxyl group (-OH), while the carbon picks up a deuterium atom.<\/p>\n\n\n\n
Because 2,3-butanedione has two ketone groups<\/strong>, each one will be reduced by NaBD4. So, both carbonyl groups are converted into hydroxyl groups, and each carbon that was originally part of a carbonyl group gains a deuterium<\/strong> atom.<\/p>\n\n\n\n
Step-by-step transformation:<\/h3>\n\n\n\n
    \n
  • Before Reaction:<\/strong> CH3\u2013CO\u2013CO\u2013CH3 (a diketone)<\/li>\n\n\n\n
  • After Reaction with NaBD4:<\/strong> Each C=O becomes C\u2013OD (from methanol as solvent, the OH forms, but the D comes from the NaBD4)<\/li>\n\n\n\n
  • The carbon atom of the C=O gets a D<\/strong> attached from the NaBD4, and the oxygen becomes an -OH group (protonated from the solvent).<\/li>\n<\/ul>\n\n\n\n
    So, the final product is:<\/p>\n\n\n\n
    CH3\u2013C(OD)\u2013C(OD)\u2013CH3<\/strong><\/p>\n\n\n\n
    But in practice, the \u2013OD is rapidly protonated by methanol to form \u2013OH, so the alcohol groups appear as \u2013OH, and the D remains at the carbon.<\/p>\n\n\n\n
    Hence, the product is:<\/p>\n\n\n\n
    CH3\u2013CH(OD)\u2013CH(OD)\u2013CH3<\/strong><\/p>\n\n\n\n
    Which corresponds to structure b<\/strong> \u2013 a molecule with two hydroxyl groups and two deuterium atoms added to the carbons that were formerly carbonyls.<\/p>\n\n\n\n
    This is why b<\/strong> is the correct answer.<\/p>\n\n\n\n
    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
    Sodium borodeuteride (NaBD4) is a reagent that has analogous reactivity to sodium borohydride. It acts as a source of deuteride (D-) instead of hydride (H-). Which of the structures below (a-f) is the product of the following reaction? NaBD4 + CH3OH OH OH OH HO- OH H2CO OCH3 The Correct Answer and Explanation is: The […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-27393","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27393","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=27393"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27393\/revisions"}],"predecessor-version":[{"id":27399,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27393\/revisions\/27399"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27393"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27393"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}