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Chemistry: Post your doubts here!
Cl and O are similar in electronegativity and cancels each other out to an extent. Free radical substitution. By keeping track of the Br, you can see that the Br is reduced (oxidation state decreased form 0 to -1) Br2 --> HBr -
Chemistry: Post your doubts here!
Adding them does not necessary give us octene, cause there might be other products that are not listed. If we stick to the thinking that these are the only products, it is still possible. Just balance the equation aC4H8 --> bC2H4 + cC3H6 + dC3H8 + eC4H6 The mistake made is that you wrongly... -
Chemistry: Post your doubts here!
Cracking of octene produces molecules that have less carbons than octene. W, X, Y and Z are thus all possible products of cracking of octene. -
Chemistry: Post your doubts here!
Under increased pressure, the total pressure of the system increases. Although the mole faction of HI does not change, the partial pressure of HI increases as partial pressure = mole fraction X total pressure -
Chemistry: Post your doubts here!
Keep in mind we are not just cracking a single molecule. Countless molecules are being broken down. Any molecule that has less carbon than the original molecule is a possible product. -
Chemistry: Post your doubts here!
Two more CHCl=CHCH2Cl (Cis-trans) -
Chemistry: Post your doubts here!
Al2O3 is insoluble in water. MgO dissolves slightly in water to form an alkaline solution. That itself is not alkaline enough to react with Al2O3. -
Chemistry: Post your doubts here!
It would be in the 2 and 6 position with respect to OH. OH is a stronger activating group compared to alkyl groups, there is also the issue of the steric hindrance of the large alkyl group. -
Chemistry: Post your doubts here!
Do you have a diagram of the compound? Or is it something you thought of? -
Chemistry: Post your doubts here!
You can try viewing it from the end on the left side. Then visualize them as clock faces, using references with H and CH3 at the 12 o'clock positions. -
Solutions to Challenging 9701 MCQ Paper 1
Based on my interactions with students in the academic forums, I have compiled a set of 20 calculation questions (some sample shown below) from past year papers that they often require help with. It is not surprising to find similar questions being repeated throughout the years, as such... -
Chemistry: Post your doubts here!
MnO4- is an oxidizing agent, in acidic conditions, it would form Mn2+. The Fe2+ would in turn be oxidized to Fe3+. E(cell) = E(reduction) - E(oxidation) > 0 for reaction to be feasible. -
Chemistry: Post your doubts here!
It is not Cl- that reacts with the alkene in the first step. Don't forget that alkenes are electron rich, they will attract ELECTROPHILES. Replace the Br-Br with Cl-Cl for the diagrams below. -
Chemistry: Post your doubts here!
This is a common misconception by A level students. For hydrogen bonding, you need a hydrogen bond acceptor and a hydrogen bond donor. For a hydrogen bond donor, you need a H joined to N, O, or F. For a hydrogen bond acceptor, you need lone pairs on N, O, or F (there is no need for it to have... -
Chemistry: Post your doubts here!
Is this an A level question and ms? Can post whole marking scheme here? -
Chemistry: Post your doubts here!
On the benzene ring, you can add 2 bromine atoms. Next to the OH groups. -
Chemistry: Post your doubts here!
Priority numbering is to the halogen. Naming order is alphabetical. -
Chemistry: Post your doubts here!
1-chloro-4-ethyl-5-methylhexane is correct. The smallest number assigned to the halogen group. Last time I mentioned that "halogen is lower priority than alkane" , I realized that can be misinterpreted. Between halogen and alkyl group, we assign a lower number to the halogen group, but we...