Ncert Solutions Chemistry Class 12th
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2 months agoContributor-Level 10
This is electrophilic substitution reaction which is determine by electronic effect of 
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2 months agoContributor-Level 10
E2 elimination. Most acidic proton is removed. Fluorine is more electronegative.
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2 months agoContributor-Level 10
Seliwanoff's test distinguishes aldoses from ketoses. Sucrose hydrolyzes to glucose (aldose) and fructose (ketose). Fructose gives a red color.
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2 months agoContributor-Level 10
gives iodoform test and slow Lucas test, so it's a methyl secondary alcohol. D gives fast Lucas test, so it's a tertiary alcohol. The Grignard products must be tertiary and secondary alcohols. So A and B must be an aldehyde and a ketone.
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2 months agoContributor-Level 10
. [XeF? ]? has 7 electron pairs (5 bonding, 2 lone), pentagonal planar. XeO? F? has 5 electron pairs (trigonal bipyramidal).
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2 months agoContributor-Level 9
Sol. (x/m) = k (P)¹/?
log (x/m) = logk + 1/n logP
Slope = 1/n = 2 So n = 1/2
Intercept ⇒ logk = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3² = 48
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2 months agoContributor-Level 10
Reaction (1) is SN1 (rate independent of [OH? ]). Reaction (2) is E2 (rate depends on [OH? ]).
Statement (B) is correct. Changing concentration of base will have no effect on reaction (1).
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2 months agoContributor-Level 9
Cesium has lowest ionisation enthalpy and hence it can show photoelectric effect to the maximum extent hence it is used in photo electric cell.
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2 months agoTaking an Exam? Selecting a College?
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