
- AIntermolecular $C - N$ coupling
- BIntramolecular $C- N$ coupling
- CIntermolecular $N - N$ coupling
- ✓Intramolecular $N -N$ coupling

Benzotriazole may be obtained preparatively ( $75$ $\%$ yield) in this way.
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$(i)$ Glucose $+ ROH \quad \stackrel{\text { dry } HCl }{\longrightarrow}$ Acetal
$\xrightarrow[{{{\left( {C{H_3}CO} \right)}_2}O}]{{x\,eq.of}}$ acetyl derivative
$(ii)$ Glucose $\xrightarrow{{Ni/{H_2}}}A\xrightarrow[{{{\left( {C{H_3}CO} \right)}_2}O}]{{y\,\,eq.\,of}}$ acetyl derivative.
$(iii)$ Glucose $\xrightarrow[{{{\left( {C{H_3}CO} \right)}_2}O}]{{z\,ed.\,of}}$ acetyl derivative.
$' x ^{\prime},{ }^{\prime} y ^{\prime}$ and ${ }^{\prime} z^{\prime}$ in these reactions are respectively.
|
List $-I$ (Element) |
List $-II$ (Electronic Configuration) | ||
| $A.$ | $N$ | $I.$ | $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^5$ |
| $B.$ | $S$ | $II.$ | $[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4$ |
| $C.$ | $Br$ | $III.$ | $[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^3$ |
| $D.$ | $Kr$ | $IV.$ | $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^6$ |
Choose the correct answer from the options given below:
Reason : Energy of the activated complex is higher than the energy of reactant molecules.

