{"id":230201,"date":"2025-06-09T05:14:51","date_gmt":"2025-06-09T05:14:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230201"},"modified":"2025-06-09T05:14:53","modified_gmt":"2025-06-09T05:14:53","slug":"complete-the-synthesis-below-by-selecting-or-drawing-the-reagents","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/09\/complete-the-synthesis-below-by-selecting-or-drawing-the-reagents\/","title":{"rendered":"Complete the synthesis below by selecting or drawing the reagents."},"content":{"rendered":"\n

Complete the synthesis below by selecting or drawing the reagents. Structure A OH MgBr 1. Reagent A 2. Reagent B Reagent 2. H3O+<\/p>\n\n\n\n

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

To complete this synthesis, we\u2019ll analyze the structure and the reagents given:<\/p>\n\n\n\n

You provided Structure A<\/strong>, which has an \u2013OH group and an MgBr<\/strong> group, indicating that it is a Grignard reagent<\/strong> formed from a halide (Br) and a carbon backbone with an \u2013OH group.<\/p>\n\n\n\n

Let\u2019s assume this is a Grignard addition reaction<\/strong>, which typically involves a carbonyl compound<\/strong> (like an aldehyde or ketone) reacting with a Grignard reagent to form an alcohol after acidic workup.<\/p>\n\n\n\n

\n\n\n\n

Proposed Synthesis:<\/strong><\/h3>\n\n\n\n

Structure A:
OH\u2013CH\u2082\u2013CH\u2082\u2013MgBr<\/strong> (a Grignard reagent with an alcohol on one end)<\/p>\n\n\n\n

We want to create this using:<\/p>\n\n\n\n

Step 1: Reagent A = Mg in dry ether<\/strong>
This forms the Grignard reagent from a bromoalkane:
Br\u2013CH\u2082\u2013CH\u2082\u2013OH + Mg \u2192 HO\u2013CH\u2082\u2013CH\u2082\u2013MgBr<\/strong><\/p>\n\n\n\n

Step 2: Reagent B = Carbonyl compound (e.g., formaldehyde or aldehyde)<\/strong>
The Grignard reagent attacks the carbonyl carbon to form a new C\u2013C bond.<\/p>\n\n\n\n

Let\u2019s say we want to form a primary alcohol<\/strong> on the terminal end. That means Reagent B = formaldehyde (HCHO)<\/strong>.<\/p>\n\n\n\n

Then, in Step 2:<\/p>\n\n\n\n

HO\u2013CH\u2082\u2013CH\u2082\u2013MgBr + HCHO \u2192 HO\u2013CH\u2082\u2013CH\u2082\u2013CH\u2082\u2013OH (after H\u2083O\u207a workup)<\/strong><\/p>\n\n\n\n

So, the complete reaction is:<\/p>\n\n\n\n

\n\n\n\n

Answer:<\/strong><\/h3>\n\n\n\n
    \n
  • Reagent A<\/strong>: Magnesium (Mg), dry ether<\/li>\n\n\n\n
  • Reagent B<\/strong>: Formaldehyde (HCHO)<\/li>\n\n\n\n
  • Final step<\/strong>: Acid workup (H\u2083O\u207a)<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

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

    This synthesis involves using a Grignard reagent<\/strong>, which is a powerful organometallic compound formed by reacting an alkyl halide with magnesium metal in dry ether. Grignard reagents are highly nucleophilic and react readily with electrophilic carbon atoms, especially those in carbonyl groups.<\/p>\n\n\n\n

    In this case, the starting compound is a bromoalcohol: Br\u2013CH\u2082\u2013CH\u2082\u2013OH<\/strong>. When treated with magnesium (Reagent A)<\/strong> in dry ether, the bromine is replaced by a magnesium bromide group, forming the Grignard reagent HO\u2013CH\u2082\u2013CH\u2082\u2013MgBr<\/strong>.<\/p>\n\n\n\n

    This Grignard reagent can then react with a carbonyl compound. The most straightforward choice is formaldehyde (HCHO)<\/strong>, a simple aldehyde. Grignard reagents attack the electrophilic carbon in the carbonyl group, leading to the formation of an alkoxide intermediate.<\/p>\n\n\n\n

    Following the reaction with formaldehyde, the addition product is an alkoxide, which upon protonation with water or dilute acid (H\u2083O\u207a), yields a primary alcohol<\/strong>. The overall transformation lengthens the carbon chain by one carbon unit and introduces an \u2013OH group at the end, creating a 1,3-diol (HO\u2013CH\u2082\u2013CH\u2082\u2013CH\u2082\u2013OH)<\/strong>.<\/p>\n\n\n\n

    Thus, by choosing Mg<\/strong> and formaldehyde<\/strong> as Reagents A and B respectively, and finishing with an acid workup, the synthesis is successfully completed. This method illustrates a classic application of Grignard reagents in carbon\u2013carbon bond formation, a fundamental concept in organic synthesis.<\/p>\n\n\n\n

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

    Complete the synthesis below by selecting or drawing the reagents. Structure A OH MgBr 1. Reagent A 2. Reagent B Reagent 2. H3O+ The Correct Answer and Explanation is: To complete this synthesis, we\u2019ll analyze the structure and the reagents given: You provided Structure A, which has an \u2013OH group and an MgBr group, indicating […]<\/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-230201","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230201","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=230201"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230201\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230201"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230201"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230201"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}