Organic Chemistry – Some Basic Principles and Techniques

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Question
CBSEENCH11008295

Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear.

  • CH3 - C ≡ C - CH3

  • CH2 = CH - CH2 - C ≡ CH

  • CH3 - CH2 - CH2 - CH3

  • CH3 - CH = CH - CH3

Solution

A.

CH3 - C ≡ C - CH3

CH3 - C ≡ C - CH3 linear, because C2 and C3 are sp hybridised.

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Question
CBSEENCH11008356

In the following the most stable conformation of n-butane is

Solution

B.

The conformation in which the heavier groups are present at maximum possible distances so that the forces of repulsion get weak is more stable. 
Among the given conformations of n- butane, the conformation is shown in option (b) ie, anti conformation is most stable as in it the bulkier groups I(ie, CH3 group) are present at the maximum possible distance. 

Question
CBSEENCH11008347

Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by

  • oxidation

  • cracking

  • distillation under reduced pressure

  • hydrolysis

Solution

B.

cracking

Lower hydrocarbon exists in a gaseous state while higher ones are in liquid state or solid state.
On cracking or pyrolysis, the hydrocarbon with higher molecular mass gives a mixture of hydrocarbons having lower molecular mass. Hence, we can say that by cracking a liquid hydrocarbon can be converted into a mixture of gaseous hydrocarbons.

Question
CBSEENCH11008263

Structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is

Solution

B.

Question
CBSEENCH11008284

Sulphur trioxide can be obtained by which of the following reaction

  • CaSO subscript 4 space plus straight C space rightwards arrow with increment on top
  • Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 space rightwards arrow with increment on top
  • straight S space plus straight H subscript 2 end subscript SO subscript 4 end subscript rightwards arrow with increment on top
  • straight H subscript 2 end subscript SO subscript 4 end subscript plus PCl subscript 5 space rightwards arrow with increment on top

Solution

B.

Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 space rightwards arrow with increment on top straight a right parenthesis space CaSO subscript 4 space plus straight C rightwards arrow with increment on top space CaO space plus SO subscript 2 space plus CO
straight b right parenthesis space Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 space rightwards arrow with increment on top Fe subscript 2 straight O subscript 3 space plus 3 SO subscript 3
straight c right parenthesis space straight S space plus straight H subscript 2 SO subscript 4 space rightwards arrow with increment on top 3 SO subscript 2 space plus 2 straight H subscript 2 straight O
straight d right parenthesis space straight H subscript 2 SO subscript 4 space plus PCl subscript 5 space rightwards arrow with increment on top space SO subscript 3 HCl space plus POCl subscript 3 space plus HCl
Thus comma space SO subscript 3 space is space obtained space by space heating space space Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3.

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