Pairs of initial and final elements in a column of datetime pandas

I have a time column like this:

and I need to get the first and last element of each subset of times that are continuous and change only every 15 minutes. I.e., the first subset would be: {"start_time": 7:00:00, "end_time":7:30:00} and the second subset would be: {"start_time":8:00:00, "end_time":8:45:00}, so I need to return a list with those two dictionaries like:

[{"start_time": 7:00:00, "end_time":7:30:00}, {"start_time":8:00:00, "end_time":8:45:00}]

another example:

Returns: [{"start_time": 7:00:00, "end_time":7:30:00}]

The last one:

returns: [{"start_time": 7:00:00, "end_time":7:30:00}, {"start_time":10:00:00, "end_time":10:30:00}]

Covvert the column to timedelta then calculate the diff between previous and current row to get the number of seconds now compare it with 5 minuts to flag the change. Then group the dataframe by continuous blocks (i.e mask.cumsum()) and aggregate Time with first and last

mask = pd.to_timedelta(df['Time']).diff().dt.total_seconds() != 900
df.groupby(mask.cumsum())['Time'].agg(['first', 'last']).to_dict(orient='records')

[{'first': '7:00:00', 'last': '7:30:00'},
 {'first': '8:00:00', 'last': '8:45:00'}]

Something like this works:

import pandas as pd

times = [
"7:00:00",
"7:15:00",
"7:30:00",
"8:00:00",
"9:00:00",
"10:00:00",
"10:15:00",
"10:30:00",
"11:00:00"]
times = [pd.Timedelta(t) for t in times]

df = pd.DataFrame(times, columns=['Times'])

fifteen = pd.Timedelta(minutes=15)
prev = None
for t in df['Times']:
    if prev:
        if t - prev == fifteen:
            prev = t
            continue
        if curr != prev:
            print({'start':curr, 'end':prev})
    curr = prev = t

Output:

{'start': Timedelta('0 days 07:00:00'), 'end': Timedelta('0 days 07:30:00')}
{'start': Timedelta('0 days 10:00:00'), 'end': Timedelta('0 days 10:30:00')}