Compound 'A' with molecular formula C₄H₉Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound 'A' only. When another optically active isomer 'B' of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.

(i) Write down the structural formula of both compounds ' A' and ' B '.

(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

This is a short answer type question as classified in NCERT Exemplar

(i) Because the rate of reaction of compound ‘A' with aqueous KOH is solely dependent on the concentration of ‘A, ' the reaction mechanism is S_N¹and ‘A' is 2-Bromo-2-methylpropane (tertiary bromide).

‘B', on the other hand, is an isomer of ‘A' and is optically active. As a result, ‘B' has to be 2-Bromobutane. Because the rate of reaction of ‘B' with aqueous KOH is determined by its concentration, the reaction mechanism is S_N².

(ii) Compound ‘B' will have an inverted conformation as a result of the S_N² reaction,

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This is a short answer type question as classified in NCERT Exemplar

(i) Because the rate of reaction of compound ‘A' with aqueous KOH is solely dependent on the concentration of ‘A, ' the reaction mechanism is S_N¹and ‘A' is 2-Bromo-2-methylpropane (tertiary bromide).

‘B', on the other hand, is an isomer of ‘A' and is optically active. As a result, ‘B' has to be 2-Bromobutane. Because the rate of reaction of ‘B' with aqueous KOH is determined by its concentration, the reaction mechanism is S_N².

(ii) Compound ‘B' will have an inverted conformation as a result of the S_N² reaction, resulting in an inverted product.

Structure ‘A’ is and structure ‘B’ is 

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<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p><strong> (i) </strong>Because the rate of reaction of compound ‘A' with aqueous&nbsp;<strong>KOH</strong> is solely dependent on the concentration of ‘A, ' the reaction mechanism is <strong>S_N¹</strong>and ‘A' is 2-Bromo-2-methylpropane (tertiary bromide).</p><p>‘B', on the other hand, is an isomer of ‘A' and is optically active. As a result, ‘B' has to be 2-Bromobutane. Because the rate of reaction of ‘B' with aqueous <strong>KOH</strong> is determined by its concentration, the reaction mechanism is <strong>S_N²</strong>.</p><p><strong> (ii) </strong>Compound ‘B' will have an inverted conformation as a result of the&nbsp;<strong>S_N² </strong>reaction, resulting in an inverted product.</p><p>Structure ‘A’ is and structure ‘B’ is&nbsp;</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1746530247phpKwQn30_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1746530247phpKwQn30.jpeg" alt="" width="490" height="110"></picture></div></div>

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