For the reaction sequence given below, the correct statement(s) is(are)

The given reaction sequence proceeds through the following steps:

Step 1: Oxidation of Tetralin to form X
When tetralin (1,2,3,4-tetrahydronaphthalene) is treated with acidic potassium permanganate (${\mathrm{KMnO}}_4/H^{+}$) under heating ($\Delta$), the non-aromatic ring undergoes oxidative cleavage to yield phthalic acid (benzene-1,2-dicarboxylic acid).
Subsequently, heating phthalic acid with ammonia (${\mathrm{NH}}_3$) results in the loss of two water molecules ($-2H_{2}O$), forming phthalamide:
$C_6{H}_4{\left(\mathrm{COOH}\right)}_2+2{\mathrm{NH}}_3\to C_6{H}_4{\left({\mathrm{CONH}}_2\right)}_2+2H_{2}O$
Thus, X is phthalamide, which contains oxygen (carbonyl groups).

Step 2: Conversion of X to Y
(i) Strong heating of phthalamide (X) results in the elimination of a molecule of ammonia (${\mathrm{NH}}_3$) to form phthalimide.
(ii) Treating phthalimide with ethanoic (alcoholic) potassium hydroxide ($\mathrm{KOH}$) deprotonates the imide nitrogen, producing potassium phthalimide.
(iii) Nucleophilic substitution of potassium phthalimide with an alkyl bromide ($\mathrm{R-Br}$) yields N-alkylphthalimide:
Thus, Y is N-alkylphthalimide, which is also an oxygen-containing compound.

Step 3: Hydrolysis of Y to form Z
Basic hydrolysis of N-alkylphthalimide (Y) using sodium hydroxide ($\mathrm{NaOH}$) yields sodium phthalate (the aromatic compound) and an aliphatic primary amine (Z, $\mathrm{R-}{\mathrm{NH}}_2$). This entire process represents the Gabriel Phthalimide Synthesis.

Therefore, both compound X (phthalamide) and compound Y (N-alkylphthalimide) contain oxygen.