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A sample of 4.50 mg of unknown alcohol is added to CH3MgBr when 1.68 mL of CH4 at STP is obtained. The unknown alcohol is:
  1. methanol
  2. ethanol
  3. 1-propanol
  4. 1-butanol
Solution (C)
All the four options given are monohydric alcohols. When a monohydric alcohol is treated with CH3MgBr, the number of moles of CH4 gas produced is equal to the number of moles of alcohol.
Description: 46418.png
If M is the molecular weight of alcohol, then
Description: 46425.png
On solving, M = 60. Therefore, the unknown alcohol is 1-propanol.


A compound (X) when passed through dil. H2SO4 containing H2SO4 gives compound (Y), which on reaction with HI and red phosphorous gives C2H6. The compound (X) is:
  1. ethane
  2. ethyne
  3. 2-butene
  4. 2-butyne
Solution (B)
The compound (X) is likely to be alkyne which reacts with water in the presence of H2SO4 and HgSO4 as catalyst to form a carboxyl compound. HI and red phosphorous can reduce a carboxyl compound to alkane having same number of carbon atoms.
Therefore (Y) is likely to be acetaldehyde, which is the hydration product of ethyne.
Description: 46432.png Description: 46440.png Description: 46448.png
Description: 46458.png
Therefore, (X) is ethyne.


When 2-methylbutane is chlorinated, the percentage of (CH3)2–CH–CH2–CH2Cl is ……… Assume reactivity ratio of 3°H : 2°H : 1°H = 5: 3.8 : 1.
  1. 28%
  2. 35%
  3. 23%
  4. 14%
Solution (D)
Chlorination of 2-methylbutane gives monochlorinated products:
Description: 46468.pngDescription: 46475.png
Description: 46483.pngDescription: 46493.png
The reactivity ratio of 3°H : 2°H : 1°H towards chlorination is 5 : 3.8 : 1.
Compound reactivity fr × Probability fr = Number of parts
Total number of parts = 21.6
Percentage of Description: 46501.png


During debromination of meso-dibromobutane, the major compound formed is:
  1. n-butane
  2. 1-butene
  3. cis-2-butene
  4. trans-2-butene
Solution (D)
In debromination reaction, we have to ensure that the two leaving groups, i.e., the two Br atoms, are 180° to each other in the same plane before they are lost. From the above conformations of meso-dibromobutane drawn on Newmann projection, it is clear that deracination results in the formation of trans-2-butene.


The addition of HCl to 3,3,3-trichloropropene gives:
  1. Cl3CCH2CH2Cl
  2. Cl3CCH(Cl)CH3
  3. Cl2CHCH(Cl)CH2C
  4. Cl2CHCH2CHCl2
Solution (A)
CCl3 is a strongly electron-withdrawing group. The addition of HCl to Description: 46647.png double bond of 3,3,3-trichloropropene does not follow the Markovnikoff’s rule because intermediate secondary carbonium ion is destabilized by the –I effect of CCl3 group.
Description: 46655.png
Instead the addition follows anti-Markovnikoff’s rule because primary carbonium ion becomes relatively more stable.
Description: 46664.pngDescription: 46674.png


The treatment of propene with Cl2 at 500–600°C produces:
  1. 1,2-dichloropropene
  2. allyl chloride
  3. 2,3-dichloropropene
  4. 1,3-dichloropropene
Solution (B)
At high temperature, Cl2 does not add to alkenes. Chlorine molecule undergoes homolytic cleavage at high temperature forming Cl radicals which initiate substitution reaction at the allylic position. Therefore, when propene reacts with Cl2 at high temperature, allyl chloride is formed.
Description: 46685.pngDescription: 46693.png


The main product produced in the dehydrohalogenation of 2-bromo-3,3-dimethylbutane is:
  1. 3,3-dimethylbut-1-ene
  2. 2,3-dimethylbut-1-ene
  3. 2,3-dimethylbut-2-ene
  4. 4-methylpent-2-ene
Solution (C)
Description: 47263.png
Description: 47273.png
The reaction follows E1 mechanism. In the first step, a secondary C+ ion is formed which rearranges to give more stable tertiary C+ ion. In the second step, the tertiary C+ ion loses H+ from its adjacent C atom to give more stable alkene, i.e., 2,3-dimethylbut-2-ene.


The intermediate during the addition of HCl to propene in the presence of peroxide would be:
  1. Description: 47099.png
  2. Description: 47109.png
  3. Description: 47118.png
  4. Description: 47125.png
Solution (B)
 of HCl to propene follows Markovnikoff’s rule because peroxide free radicals are unable to break H–Cl bond due to its high bond dissociation energy.
Description: 47133.pngDescription: 47164.png
The reaction proceeds via ionic mechanism with Description: 47173.png as intermediate.


The product of the reaction Description: 47181.png would be:
  1. Description: 47188.png
  2. Description: 47195.png
  3. Description: 47206.png
  4. H2C=OD–CH2CHO
Solution (B)
Description: 47214.png
Description: 47223.png removes H+ from HC ≡ CD and not D+ because C–H bond energy is lower than C–D. The anion attacks the C atom of carbonyl group.


2-Methylpent-2-ene on ozonolysis will give:
  1. propanal only
  2. propanal and ethanol
  3. propanone and propanal
  4. propan-2-ol and ethanol
Solution (C)
The products of the above reaction are propanone and propanal.

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