Multiple Choice QuestionsThe IUPAC name of (CH3)3 C-CH=CH2 is
1, 1, 1-trimethyl-2-propene
3, 3, 3-trimethyl-2-propene
2, 2-dimethyl-3-butene
3, 3-dimethyl-1-butene
An organic compound answers Molisch's test as well as Benedict's test. But it does not answer Scliwanoffs test. Most probably, it is
sucrose
protein
fructose
maltose
In qualitative analysis, in order to detect second group basic radical, H2S gas is passed in the presence of dilute HCl to
increase the dissociation of H2S
decrease the dissociation of salt solution
decrease the dissociation of H2S
increase the dissociation of salt solution
Pick out the unsaturated fatty acid from the following
stearic acid
Lauric acid
oleic acid
Palmitic acid
The first fraction obtained during the fractionation of petroleum is
hydrocarbon gases
kerosene oil
gasoline
diesel oil
During the fusion of an organic compound with sodium metal, nitrogen of the compound is converted into :
NaNO2
NaNH2
NaCN
NaNC
Ethyl chloride on heating with AgCN forms a compound X. The functional isomer of X is
C2H5NC
C2H5NH2
C2H5CN
None of these
20 mL of 0.5 N HCl and 35 mL of 0.1 N NaOH are mixed. The resulting solution will
be neutra
be basic
turn phenolphthalein solution pink
turn methyl orange red
C.
turn phenolphthalein solution pink
(i) 20 mL of 0.5 N HCl
0.5 N - 1000 mL 0.5 mol HCl is present in 20 mL
= = 1.0 × 10-2
(ii) 35 mL of 0.1 N NaOH
0.1 N - 1000 mL of 0.1 mol of NaOH
NaOH present in 35 mL =
Total volume = 20 + 35 = 55 mL
(Resulting mole in solution)
(1.0 - 0.35) 10-2 = 0.65 × 10-2 mol HCl
HCl = H+ + Cl-
[HCl] = [H+] + [Cl-]
55 mL contains 0.65 × 10-2 mole of H+ ions
1000 mL =
pH = -log[H+] = -log (6.5 / 55)
= log 55 - log 6.5 = 0.92
Due to acidic nature of solution the colour of phenolphthalein becomes pink.
H3C-C(Cl) = CH - CH(CH3) - CH3
2-chloro-4-methyl-2pentene
4-chloro-2-methyl-3-pentene
4-methyl-2-chloro-2-pentene
2-chloro-4,4-dimethyl-2-butene
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