General Properties of Hydrocarbons - Free MCQ Practice Test with solutions,

Among all of them hydrogen 'b' will be most leachive towards chlorination as it is 2^o hydrogen and it will get less electronegativity of fluorine atom and 'd' will be least stable as it is adjacent to the fluorine atom and does not have 3^o - 2^o hydrogen.

Bromine is more selective
∴ it will form 3^o halide if there is presence of 3^o hydrogen.
∴ Q, S, T from 3^o halide as major and P, R, U form 2^o halide as major.

Acetophenone can be prepared from benzene by its electrophilic substitution (Friedel-Craft acylation) reaction as follows

The correct order of increasing reactivity towards HCL is

As the number of alkyl groups attached to the carbon atoms increases, the + I effect of the alkyl groups increases and the alkene becomes electron rich. Hence, the reactivity towards HCl increases as pi electrons of the double bond are donated to proton.
Also as the stability of alkene decreases (cis isomer is less stable than trans isomer), the reactivity towards acid catalyzed hydration increases. Therefore, Option C is correct answer for this.

Br^o is less reactive and more selective and so the most stable free radical (3^o) will be the major product.

Hexane on single monochlorination gives 3-structural and 2-stereo isomers i.e., total 5 chloro derivatives as follows

Temperature is the effective factor for dehydration of alcohols by Al₂O₃.

At 220^o - 250^o it forms ether.

Alkyl halides react with sodamide in an elimination to yield an alkene or alkyne (from dihalides). For instance,

1, 2 - dibromo ethane acetylene
Hence, product has both sp-hybridised carbon.

Two monochlorobutanes (structural isomers) will be obtained on chlorination of n-butane.

one monochlorintaion product.
From, the given reaction we can say that this 2, 2, 3, 3 tetramethyl butane give only one chlorination product as chlorine is added to one of the -CH₃ group.

In this reaction, 1^o carbonium ion is formed which rearranges to form 3^o carbonium ion. Hence, 2-methyl-2-butene is formed as the main product.

Br₂ addition is anti-addition.
Now remember 'CAR' and 'TAM' rule.
C Cis T Trans
A Anti A Anti
R Racemic M Meso

tep 1: Calculate the number of moles of carbon and hydrogen

• The molar mass of carbon is 12 g/mol, and the molar mass of hydrogen is 1 g/mol.
• Number of moles of carbon = 10.5 g / 12 g/mol = 0.875 mol
• Number of moles of hydrogen = 1 g / 1 g/mol = 1 mol

Step 2: Find the molar ratio of carbon to hydrogen

• The molar ratio of carbon to hydrogen can be calculated by dividing the number of moles of carbon by the number of moles of hydrogen.
• Molar ratio = 0.875 mol / 1 mol = 0.875

Step 3: Determine the empirical formula of the hydrocarbon

• The empirical formula of the hydrocarbon can be determined by dividing the molar ratio of carbon to hydrogen by the smallest whole number.
• In this case, the empirical formula would be CH

Step 4: Calculate the molar mass of the empirical formula

• The molar mass of CH = 12 g/mol (carbon) + 1 g/mol (hydrogen) = 13 g/mol

Step 5: Calculate the number of moles of the hydrocarbon

• Number of moles = 2.8 g / 13 g/mol = 0.2154 mol

Step 6: Determine the molecular formula of the hydrocarbon

• The molecular formula of the hydrocarbon can be determined by dividing the molar mass of the hydrocarbon by the molar mass of the empirical formula.
• Molecular formula = 0.2154 mol / 0.875 mol = 0.2461
• Since the molecular formula must be a whole number, the closest whole number to 0.2461 is 0.25.
• Therefore, the molecular formula of the hydrocarbon is C7H8, which means the number of carbon in the molecular formula is 7.

Therefore, the correct answer is C: 7.

(I)
The mixture of is called sodalime when it reacts with sodium salt of carboxylic acid, it gives corresponding alkane. The reaction is called decarboxylation of salts of carboxylic acids.
(II)
The above reaction is called Clemmensen-reduction. It converts aldehydes and ketones to corresponding alkane.
(III)
This reaction is used for reduction of alkynes which gives corresponding alkenes.
This reaction is used for reduction of alkynes which gives corresponding alkenes.
(iv)
Hence, option (3) is the correct answer.

Electrolysis of sodium or potassium salt of carboxylic acid gives good yield of hydrocarbon

At anode

At cathode

More is the stability, lower is the heat of combustion (See Baeyer's strain theory). Cyclohexane is more stable.

Free radicals are formed as intermediates.