Organic Chemistry

Step 1: Analyzing the Molecular Formula:
The given information states the compound has 3 Carbon atoms and 6 Hydrogen atoms. The molecular formula is C_3H_6.

Step 2: Matching with General Formulas:
Alkanes: C_nH_2n+2 (For n=3, H=8) Propane Alkenes: C_nH_2n (For n=3, H=6) Propene Alkynes: C_nH_2n-2 (For n=3, H=4) Propyne

Step 3: Verifying Unsaturation:
Unsaturated hydrocarbons contain double or triple bonds. Propene (C_3H_6) contains one carbon-carbon double bond, making it unsaturated.

Step 4: Final Answer:
The compound is propene.

Step 1: Understanding the Reaction (Cracking):
The reaction described is thermal cracking, where a long-chain saturated hydrocarbon (alkane) is broken down into a smaller alkane and an unsaturated hydrocarbon (alkene).

Step 2: Identifying the Compounds:
X (Butane): (C_4H_10). This is an alkane. All carbon-carbon bonds are single bonds (Saturated). Y (Propene): (C_3H_6). This is an alkene. It contains at least one carbon-carbon double bond (Unsaturated). Z (Methane): (CH_4). This is the simplest alkane (Saturated).

Step 3: Defining Unsaturated:
Unsaturated hydrocarbons are those that contain double or triple covalent bonds between carbon atoms. In this reaction, only propene (Y) fits this definition.
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Step 4: Final Answer:
Compound Y only is unsaturated.

Step 1: Understanding the Reaction Type:
Ethane is a saturated hydrocarbon (alkane). Alkanes undergo substitution reactions with halogens where hydrogen atoms are replaced one by one.

Step 2: The Role of UV Light:
UV light provides the energy required to break the Cl-Cl bond to start the free-radical chain reaction.

Step 3: Writing the Products:
One hydrogen atom from ethane (C_2H_6) is replaced by one chlorine atom to form chloroethane (C_2H_5Cl). The displaced hydrogen atom bonds with the second chlorine atom to form hydrogen chloride (HCl).

Step 4: Final Answer:
The balanced equation is C_2H_6 + Cl_2 C_2H_5Cl + HCl.

(a) Observation: A white precipitate of Calcium hydroxide [Ca(OH)_2] is formed, which is sparingly soluble in excess of sodium hydroxide.

(b) Observation: A gas is evolved which has a characteristic foul smell of rotten eggs (Hydrogen Sulphide, H_2S).

(c) Observation: The metal dissolves with brisk effervescence as a colorless, odorless gas (Hydrogen) is evolved, which burns with a 'pop' sound.

(a) Basicity of HCl: Hydrochloric acid (HCl) is a monobasic acid. It contains only one ionizable hydrogen atom per molecule (HCl H^+ + Cl^-). Acid salts are only formed by polybasic acids (like H_2SO_4) where there is at least one replaceable hydrogen left after partial neutralization. Since HCl has only one hydrogen, it can only form normal salts.

(b) Nuclear Charge: As we move across a period from left to right, the nuclear charge (number of protons) increases while the number of shells remains the same. This increases the force of attraction exerted by the nucleus on the shared pair of electrons, thereby increasing electronegativity.

(a) Addition: Since the hydrocarbon is unsaturated (contains multiple bonds) and reacts with hydrogen to become saturated without losing any atoms, it is an addition reaction.

(b) Catalyst: Suitable catalysts include Nickel (Ni), Platinum (Pt), or Palladium (Pd).

(c) Balanced Chemical Equation: Since the hydrocarbon has 2 carbon atoms and reacts with two moles of H_2, the starting material must be Ethyne (C_2H_2). [ C_2H_2 + 2H_2 xrightarrowNi, 200^circtextC C_2H_6 ] (Ethyne reacts with two moles of Hydrogen to form Ethane).

Based on the descriptions: P is an Alkali Metal (Group 1), Q is likely a Group 14 element (like Carbon or Silicon), and R is a Halogen (Group 17).
(a) Which element would be most difficult to reduce?
Element P. P is highly electropositive and wants to lose electrons (oxidize). Forcing it to gain electrons (reduction) is extremely difficult.
(b) Smallest atomic radius?
Element R. Atomic radius decreases across a period from left to right due to increased nuclear charge. R is furthest to the right.
(c) Decreasing order of ionization potential:
Ionization potential increases across a period. Therefore, the order of decreasing IP is:
R > Q > P

(a) Dehydrohalogenation of ethylene dibromide: [ C_2H_4Br_2 + 2KOH (textalc.) xrightarrowDelta C_2H_2 + 2KBr + 2H_2O ] The product is Ethyne (Acetylene).

(b) Reaction of excess ammonia with chlorine: [ 8NH_3 + 3Cl_2 6NH_4Cl + N_2 ] The products are Ammonium chloride (white fumes) and Nitrogen gas. (c) Dehydration of Cane Sugar: [ C_12H_22O_11 xrightarrowtextconc. H_2SO_4 12C + 11H_2O ] The concentrated sulphuric acid removes water from the sugar, leaving behind a black spongy mass of Sugar Charcoal.

1. Butanal: Butanal is an aldehyde with 4 carbon atoms. The functional group is -CHO. [ CH_3-CH_2-CH_2-CHO ]

2. Pent-2-yne: Pent-2-yne is an alkyne with 5 carbon atoms and a triple bond starting at the second carbon atom. [ CH_3-Cequiv C-CH_2-CH_3 ]

3. Isomer of n-butane (Iso-butane / 2-Methylpropane): n-butane is a straight chain of 4 carbons. Its only structural isomer is a branched chain where 3 carbons form the main chain and 1 carbon is a methyl branch. [ CH_3-CH(CH_3)-CH_3 ]

1. Propan-1-ol: The longest chain contains 3 carbon atoms (propane). The functional group is an alcohol (-OH), which gets the suffix "-ol". Since it is attached to the first carbon, it is propan-1-ol.

2. But-1-ene: The parent chain has 4 carbon atoms (but-). The presence of a double bond indicates an alkene (suffix "-ene"). The double bond starts at the first carbon, hence but-1-ene.