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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Carboxylic Acid Derivatives

NTA Abhyas 2020 · Chemistry · Carboxylic Acid Derivatives

O | | C - NH_ 2 P _ 2 O _ 5 → A (i)CH_ 3 MgBr(ii) H _ 3 O^ + → B...

O | | C - NH_ 2 P _ 2 O _ 5 → A (i)CH_ 3 MgBr(ii) H _ 3 O^ + → B (i) I _ 2 + Ca ( OH ) _ 2 (ii) H _ 2 O → C; Product is \overset{O}{\overset{| |}{C}} - {\text{NH}_{2} \overset{{P}_{2} {O}_{5}}{→} \text{A}\overset{{\text{(i)CH}_{3} \text{MgBr}\left(\text{ii}\right){H}_{3} O^{+}}{→} \text{B}\overset{\left(\text{i}\right){I}_{2} + \text{Ca}{( \text{OH} )}_{2} \left(\text{ii}\right){H}_{2} \text{O}}{→} \text{C}; Product is

Options

  1. AIdentify X and Y in the following reaction BCl _ 3 + NH _ 4 Cl 140 ^ o C/ C _ 6 H _ 5 Cl → X NaBH _ 4 → Y C∣∣O−CH3\overset{O}{\overset{| |}{C}} - C{H}_{3}
  2. BIdentify X and Y in the following reaction BCl _ 3 + NH _ 4 Cl 140 ^ o C/ C _ 6 H _ 5 Cl → X NaBH _ 4 → Y C∣∣O−OH\overset{O}{\overset{| |}{C}} - OH
  3. CIdentify X and Y in the following reaction BCl _ 3 + NH _ 4 Cl 140 ^ o C/ C _ 6 H _ 5 Cl → X NaBH _ 4 → Y C∣∣O\mathop{\mathop{C}_{| |}}_{O} Identify X and Y in the following reaction BCl _ 3 + NH _ 4 Cl 140 ^ o C/ C _ 6 H _ 5 Cl → X NaBH _ 4 → Y
  4. DIdentify X and Y in the following reaction BCl _ 3 + NH _ 4 Cl 140 ^ o C/ C _ 6 H _ 5 Cl → X NaBH _ 4 → Y C∣∣O−CH2CH3\mathop{\mathop{C}_{| |}}_{O} - C{H}_{2}C{H}_{3}

Correct answer

B. Identify X and Y in the following reaction BCl _ 3 + NH _ 4 Cl 140 ^ o C/ C _ 6 H _ 5 Cl → X NaBH _ 4 → Y C∣∣O−OH\overset{O}{\overset{| |}{C}} - OH

Step-by-step solution

O | | C - NH_ 2 P _ 2 O _ 5 → A (i)CH_ 3 MgBr(ii) H _ 3 O^ + → B (i) I _ 2 + Ca ( OH ) _ 2 (ii) H _ 2 O → C; Product is C∣∣O−NH2→P2O5\overset{O}{\overset{| |}{C}} - N{H}_{2}\overset{{P}_{2}{O}_{5}}{→} O | | C - NH_ 2 P _ 2 O _ 5 → A (i)CH_ 3 MgBr(ii) H _ 3 O^ + → B (i) I _ 2 + Ca ( OH ) _ 2 (ii) H _ 2 O → C; Product is C ≡ N\overset{\left(i\right)\text{C}{\text{H}_{3}MgBr\left(ii\right){\text{H}_{3}O^{+}}{→} O | | C - NH_ 2 P _ 2 O _ 5 → A (i)CH_ 3 MgBr(ii) H _ 3 O^ + → B (i) I _ 2 + Ca ( OH ) _ 2 (ii) H _ 2 O → C; Product is C∣∣O−CH3→(i)I2+Ca(OH)2(ii)H2O\overset{O}{\overset{| |}{C}} - {CH}_{3} \overset{\left(\text{i}\right){I}_{2} + \text{Ca}{\left(\text{OH}\right)}_{2} \left(\text{ii}\right){H}_{2} \text{O}}{→} O | | C - NH_ 2 P _ 2 O _ 5 → A (i)CH_ 3 MgBr(ii) H _ 3 O^ + → B (i) I _ 2 + Ca ( OH ) _ 2 (ii) H _ 2 O → C; Product is C∣∣O−OH\overset{O}{\overset{| |}{C}} - OH
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