C2H5Cl+NaDry ether−−−−−−→X+NaCl Number of possible structural and stereo isomers for mono chloro derivatives of 'X' are
2
3
4
5
Solution :
C4H10 has 2 structural and one chiral centre so totally 3 isomers.
Answer is B
Solution :
Let us inspect option A
Above products on oxidation yield
O O O O
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CH3−C−HO,CH3−C−C−OH,CH3−C−CH3 (unaffected)
Let us also investigate (b)
C(CH3)2
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CH3−C−CH=CH−CH3
Above products on oxidation yield,
O O O O
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CH3−C−CH3 (unaffected), CH3−C−C−OH and CH3−C−OH
So, both options (a) and (b) are correct.
CO2
HCHO
H2O
CH3OH
Solution :
This is one of the examples of combustion of alkanes.
Alkanes readily burn in excess of air or oxyge to give CO2 and H2O
i. CH4+2O2→CO2+2H2O
ii. C3H8+5O2→3CO2+4H2O
iii. CnH2n+2+(3n+12)O2△→nCO2+(n+1)H2O+Heat (Answer is option A and C)
Questin:What is the chief product obtained when n-butane is treated with bromine in the presence of light at 130∘C
Solution :
Remember Bromination of alkane.
CH3−CH2−CH2−CH3+Br2hν→
2-Bromobutane is the main product because 2∘ carbonium ion is more stable than 1∘.
answer is A
Question:
H3C≡CHNaNH2−−−−−→XCH2CH2Br−−−−−−−→Y. Identify the product 'Y' is
Pent - 2 - ene
Pent - 2 - yne
Solution :
X is CH3−C≡C−Na+
Y is CH3−C≡C−CH2−CH3
The first step removes the terminal hydrogen from the given substrate (reactant on the left). Such a removable hydrogen is called an acidic hydrogen. In the next step, the nucleophile - which is rich in electrons - attacks the C−Br bond of ethyl bromide. Because of the difference in electronegativity between carbon and bromine, this C−Br bond is polar - meaning, the bonding electrons are closer to the more electronegative bromine. This gives rise to the possibility of the carbon of the C−Br bond having a slight deficit of electrons; let's just say that this carbon misses electrons and that is where the electron-rich terminal carbon of CH3−C≡C− comes in. Hence, the reaction.
Pent - 1 - yne
Pentane
