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Which alkyne yields butanoic acid $\left({\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CO}}_2\mathrm{H}\right)$ as the only organic product on treatment with ozone followed by the hydrolysis?
1. 1-Butyne
2. 4-Octyne
3. 1-Pentyne
4. 2-Hexyne

 
(1)
(2) 
(3)
 
(4)  ${\mathrm{CH}}_3-\underset{\underset{{\mathrm{CH}}_3}{\mathrm{l}}}{\mathrm{CH}}-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2$

(1) 
(2) 
(3) 
(4) 

Which of the following reactants is suitable for preparation of methane and ethane by using one step only?
 
(1) H${}_2$C=CH${}_2$​
(2) CH${}_3$​OH
(3) CH${}_3$​−Br
(4) CH${}_3$−CH${}_2$−OH

What is the product obtained by heating the following allylic ether of phenol ?
                                            
(1)
(2)
(3)
(4)

The change in hybridization of carbon when Ethane is subjected to combustion process is-
(1) $s{p}^2$ to $s{p}^3$
(2) $s{p}^3$ to sp
(3) sp to $s{p}^3$
(4) $s{p}^2$ to $s{p}^2$

Propene ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2$ can be converted into 1-propanol by oxidation. Indicate which of reagents amongst the following is ideal to effect the above conversion?
1. ${\mathrm{KMnO}}_4\left(\mathrm{alkaline}\right)$
2. $\mathrm{Osmium} \mathrm{tetroxide}({\mathrm{OsO}}_4/{\mathrm{CH}}_2{\mathrm{CI}}_2)$
3. ${\mathrm{B}}_2{\mathrm{H}}_6 \mathrm{and} \mathrm{alk}. {\mathrm{H}}_2{\mathrm{O}}_2$
4. ${\mathrm{O}}_3/\mathrm{Zn}$

An unknown compound (A) has a molecular formula ${\mathrm{C}}_4{\mathrm{H}}_6$. When (A) is treated with excess of ${\mathrm{Br}}_2$ a new substance (B) with formula ${\mathrm{C}}_4{\mathrm{H}}_6$${\mathrm{Br}}_2$ is formed. (A) forms a white ppt. ammonical silver nitrate solution. (A) is :
1. But-1-yne
2. But-2-yne
3. But-1-ene
4. But-2-ene

$\stackrel{{\mathrm{CH}}_3-\mathrm{NH}-{\mathrm{CH}}_3}{\to }\left(A\right)\underset{\mathrm{HCl}}{\overset{\mathrm{Fe}}{\to }}\left(B\right)$; Product (B) of this reaction is :
 
(1)  
(2)  
(3)  
 
(4)  None of these
 
 

$\mathrm{ph}-{\mathrm{co}}_2\mathrm{H}\stackrel{{\mathrm{SCOI}}_2}{\to }\mathrm{A}\stackrel{{\mathrm{Me}}_2\mathrm{NH}}{\to }\mathrm{B}\stackrel{\left(\mathrm{C}\right)}{\to }\mathrm{ph}-\mathrm{CHO},$ unknown reagent (C) is:
1. ${\mathrm{LiAlH}}_4$
2. ${\mathrm{NaBH}}_4$
3. ${\mathrm{LiAlH}}_4{(\mathrm{t}-\mathrm{BuO})}_3$
4. $\mathrm{PCC}/{\mathrm{CH}}_2{\mathrm{Cl}}_2$

$\stackrel{\mathrm{MCPBA}}{\to }\left(A\right)$ ; MCPBA ⟶ metachloroperbenzoic acid Product $$(A) of the above reaction is :
(1)
(2)
(3)
(4)

Product (C) of the above reaction is :
(1) 1,3-Hexadiene
(2) 1,4-Pentadiene
(3) 1,3-Butadiene
(4) 1,3-Heptadiene

Br−(CH${}_2$​)${}_{12}$​−C=CH$\stackrel{{\mathrm{NaNH}}_2}{\to }$​(A) $\underset{\mathrm{Catalyst}}{\overset{\mathrm{Lindlar}}{\to }}$(B); Product (B) is :
(1)
(2)
(3)
 
(4)$\mathrm{Br}-\left({\mathrm{CH}}_2\right)-\mathrm{CH}={\mathrm{CH}}_2$

Friedel-Crafts acylation reaction can be used to obtain the compounds
(1)  II, III and IV
(2)  I, III and IV
(3)  I and II
(4)  II and III

$\underset{{\mathrm{H}}_2\mathrm{O}}{\overset{\mathrm{HCHO}, }{\to }}$the major product of this reaction, is:
(1)
(2)
(3)
(4)

Which of the following compounds is the slowest to react with nitrosonium ion (NO${}^{+}$)?
(1)
(2)
(3)
(4)

An organic compound ${\mathrm{C}}_4{\mathrm{H}}_6$​ on ozonolysis gives HCHO, CH₃CHO and CO₂ . The compound will be:
(1) ${\mathrm{H}}_2\mathrm{O}=\mathrm{CH}-\mathrm{CH}={\mathrm{CH}}_2$
(2) ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{C}={\mathrm{CH}}_2$​
(3) $$${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3$
(4) 

The electrophilic aromatic substitution proceeds through a:
1. Free radical mechanism
2. Sigma complex mechanism
3. Benzyne mechanism
4. Carbene mechanism

$\underset{∆}{\overset{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}\left(A\right)\underset{\left(2\right) alc.KOH}{\overset{\left(1\right)\mathrm{HBS}}{\to }}\left(B\right)\stackrel{RC{O}_{2}H}{\to }\left(C\right) .$ Product (C) is :
(1)
(2) 
(3) 
(4) 

Di-imide (${\mathrm{N}}_2{\mathrm{H}}_4$​) is used to reduce double bond of:
1. $$$-\underset{\mathrm{l}}{\mathrm{C}}=\mathrm{O}$
2. $$$-\mathrm{C}\equiv \mathrm{N}$
3. $$$-{\mathrm{NO}}_2$​
4. $$$-\mathrm{CH}=\mathrm{CH}-$

$+ {\mathrm{CBrCl}}_3\stackrel{\mathrm{hv}}{\to }$​(A) (no ring substitution) Product (A) is :
(1) $\mathrm{Ph}-{\mathrm{CH}}_2-\mathrm{Cl}$
(2) $\mathrm{Ph}-{\mathrm{CH}}_2-\mathrm{Br}$
(3) $\mathrm{Ph}-{\mathrm{CH}}_2-{\mathrm{CCl}}_3$​
(4) $\mathrm{Ph}-{\mathrm{CH}}_2-{\mathrm{CBrCl}}_2$

Which of the following bromides is the major product of the reaction shown below, assuming that there are no carbocation rearrangement?

1. 
2. 
3. 
4.  
 

 will abstract which of the hydrogen most readily?
(1) A
(2) B
(3) C
(4) D

$\underset{2.{\mathrm{NaBH}}_4}{\overset{1.\mathrm{Hg}{\left(\mathrm{OAc}\right)}_2,{\mathrm{H}}_2\mathrm{O}}{\to }}$ Product; Product is :
 
(1)
(2)
(3)
(4)