The incorrect method for the preparation of Me₃COMe in good yield is

1. Me₃CCl + MeONa →

2. \(Me_{2}C=CH_{2}\xrightarrow[(ii)NaBH_{4}]{(i) Hg(OAc)_{2}/MeOH}\)

3. Me_3CONa + MeCl ⟶

4. \(Me_{2}C=CH_{2}\xrightarrow[MeOH]{H^{+}}\)

 Product of products

1.  R - COOH only

2.  R - CHO only

3.  R - CHO and R - COOH

4.  $O O$
$\left|\right| \left|\right|$
$R -C - C - R$

$\mathrm{PH} - \mathrm{CHO} \underset{{\mathrm{CH}}_3\mathrm{COONa}/{\mathrm{H}}_2\mathrm{O}}{\overset{{\left({\mathrm{CH}}_3 - \mathrm{CO}\right)}_2\mathrm{O}}{\to }}\left(\mathrm{A}\right)$
$\mathrm{The} \mathrm{prodct} \mathrm{A} \mathrm{is}$

1.        2.   

3.          4.  

${\mathrm{C}}_2{\mathrm{H}}_4\stackrel{{\mathrm{O}}_3/\mathrm{Zn}/{\mathrm{H}}_2{\mathrm{O}}_2}{\to } \mathrm{Product} \left(\mathrm{A}\right)$

Product (A) can give

1. Haloform test

2. Aldol condensation

3. Cannizaro reaction

4. Both (A) & (B)

Consider the following reaction,



The B and C are respectively:

1.  RCHO and HCHO

2.  RCOOH and HCOOH

3.   RCH₂OH and CH₃OH

4. 

${\mathrm{CH}}_3{\mathrm{NO}}_2\stackrel{\left(3{{\mathrm{Cl}}_{2_{}}}_{}+ 3\mathrm{NaOH}\right)}{\to } \mathrm{P}. \text{'}\mathrm{P}\text{'} \mathrm{is}$

1. ${\mathrm{CCl}}_3\mathrm{CHO}$

2. ${\mathrm{CCl}}_3{\mathrm{NO}}_2$

3. ${\mathrm{CHCl}}_3$

4. ${\mathrm{CHCl}}_2{\mathrm{NO}}_2$

The ${\mathrm{pK}}_{\mathrm{a}}$ values of some of the acids are given below:

Acid	\(\mathrm{pK_a}\)	Acid	\(\mathrm{pK_a}\)
	-0.6	HI	-10.0
	4.8	\(\mathrm{HC} \equiv \mathrm{CH}\)	25
\(\overset{+}{N}H_4\)	9.4	\(H_2S\)	7.0

The correct order of leaving tendency of their conjugate bases is:

1.
2.
3.
4.

Among the following esters. form a pair of least reactive and most reactive ester towards hydrolysis in that order

1.  V. III

2.  VI, II

3.  I, IV

4.  I, V

The product which cannot be formed is:

1.
2.
3.
4.

The product formed in this reaction is

1.          2.  

3.           4.